The Exponential Function

A LevelAQAEdexcelOCR

The Exponential Function

We have met exponential functions before, but there is one specific exponential function that has special properties, and it is based around a special number: $color{red}e$ .

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e

The exponential function is $color{red}e^{x}$ .

$color{red}e=2.71828…$ is a number. It is a decimal that goes on forever
(like $pi$ ).

$color{red}e^{x}$ has special properties, most notable being that the gradient of $color{red}e^{x}$ is $color{red}e^{x}$ . This will be very important in the differentiation section of the course.

There are some key facts to remember about the graph of $y=e^{x}$ :

It crosses the $y$ -axis at $(0,1)$
As $xrightarrowinfty$ , $color{red}e^{x}color{grey}rightarrowinfty$ and as $xrightarrow -infty$ , $color{red}e^{x}color{grey}rightarrow0$
$color{red}e^{x}$ is never negative.

$y=e^{ax+b} + c$ is a transformation of $y = e^x$ , where $a$ is a horizontal stretch, $b$ moves it horizontally and $c$ moves it vertically.

$y = e^{-x}$ reflects $y=e^x$ in the $y$ -axis.

Natural Logarithm

The inverse function of $color{red}e^{x}$ is the natural logarithm $color{blue}ln(x)$ . This is the logarithm with base $color{red}e$ $(text{log}_e (x))$ .

All the laws of logarithms can be applied to the natural logarithm.

$color{blue}ln(a)color{grey}+color{blue}ln(b)color{grey}=color{blue}ln(ab)$

$color{blue}ln(a)color{grey}-color{blue}ln(b)color{grey}=color{blue}lnleft(dfrac{a}{b}right)$

$color{blue}ln(a^{b})color{grey}=color{blue}bln(a)$

The graph of the natural logarithm (in blue) is the reflection in the line $y=x$ of the graph of the exponential function (in red).

There are key facts to remember about the graph of $color{blue}y=ln(x)$ :

It crosses the $x$ -axis at $(1,0)$
As $xrightarrowinfty$ , $color{blue}ln(x)color{grey}rightarrowinfty$ and as $xrightarrow0$ , $color{blue}ln(x)color{grey}rightarrow -infty$
$color{blue}ln(x)$ does not take any values for $xleq0$

Since $ln (x)$ is the inverse of $e^x$ and is a logarithmic function, we have these formulas relating the two:

$textcolor{red}{e}^{textcolor{blue}{ln (x)}} = x$

$textcolor{blue}{ln} textcolor{red}{(e^x)} = x$

Example 1: Equations Involving the Exponential Function

Solve for $x$ :

$color{red}e^{3x}color{grey}=10$

[2 marks]

$color{red}e^{3x}color{grey}=10$

$3x=color{blue}ln(10)$

$begin{aligned}x&=dfrac{color{blue}{ln(10)}}{3}[1.2em]&=0.768end{aligned}$

Example 2: Equations Involving Logarithms

Solve for $x$ :

$color{blue}ln(4x+3)color{grey}=2$

[2 marks]

$color{blue}ln(4x+3)color{grey}=2$

$4x+3=color{red}{e^{2}}$

$4x=color{red}e^{2}color{grey}-3$

$begin{aligned}x&=dfrac{1}{4}(color{red}e^{2}color{grey}-3)[1.2em]&=1.10end{aligned}$

The Exponential Function Example Questions

Question 1: Solve for $x$ :

a) $e^{x}=2$

a) $e^{5x}=19$

c) $e^{12x}=234$

[6 marks]

a) $e^{x}=2$

$begin{aligned}x&=ln(2)[1.2em]&=0.693end{aligned}$

b) $e^{5x}=19$

$5x=ln(19)$

$begin{aligned}x&=dfrac{1}{5}ln(19)[1.2em]&=0.589end{aligned}$

c) $e^{12x}=234$

$12x=ln(234)$

$begin{aligned}x&=dfrac{1}{12}ln(234)[1.2em]&=0.455end{aligned}$

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Question 2: Solve for $x$ :

a) $ln(x+1)=4$

a) $ln(3x+2)=1.5$

a) $ln(9x+36)=0.6$

[6 marks]

a) $ln(x+1)=4$

$x+1=e^{4}$

$begin{aligned}x&=e^{4}-1[1.2em]&=53.6end{aligned}$

b) $ln(3x+2)=1.5$

$3x+2=e^{1.5}$

$3x=e^{1.5}-2$

$begin{aligned}x&=dfrac{1}{3}(e^{1.5}-2)[1.2em]&=0.827end{aligned}$

c) $ln(9x+36)=0.6$

$9x+36=e^{0.6}$

$9x=e^{0.6}-36$

$begin{aligned}x&=dfrac{1}{9}(e^{0.6}-36)[1.2em]&=-3.80end{aligned}$

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Question 3: Solve for $x$ :

$e^{2x}-13e^{x}+36=0$

[4 marks]

$e^{2x}-13e^{x}+36=0$

Note that $e^{2x}=(e^{x})^{2}$

$(e^{x})^{2}-13e^{x}+36=0$

Substitute: $y=e^{x}$

$y^{2}-13y+36=0$

$(y-9)(y-4)=0$

$y=9$ or $y=4$

Reverse substitution:

$e^{x}=9$ or $e^{x}=4$

$x=ln(9)$ or $x=ln(4)$

$x=2.20$ or $x=1.39$

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Question 4: Solve for $x$ :

$ln(4x+3)-2ln(x)=5$

[5 marks]

$ln(4x+3)-2ln(x)=5$

$ln(4x+3)-ln(x^{2})=5$

$lnleft(dfrac{4x+3}{x^{2}}right)=5$

$dfrac{4x+3}{x^{2}}=e^{5}$

$4x+3=e^{5}x^{2}$

$e^{5}x^{2}-4x-3=0$

Use quadratic formula:

$begin{aligned}x&=dfrac{-bpmsqrt{b^{2}-4ac}}{2a}[1.2em]&=dfrac{4pmsqrt{(-4)^{2}-4times e^{5}times(-3)}}{2e^{5}}[1.2em]&=dfrac{4pmsqrt{16+12e^{5}}}{2e^{5}}[1.2em]&=dfrac{2pmsqrt{4+3e^{5}}}{e^{5}}end{aligned}$

$x=0.156$ or $x=-0.129$

We can discount the negative solution because $ln(x)$ is not valid for negative $x$ .

$x=0.156$

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Additional Resources

Exam Tips Cheat Sheet

A Level

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A Level

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Specification Points Covered

F1 – Know and use the function $a^x$ and its graph, where $a$ is positive
Know and use the function $e^x$ and its graph
F2 – Know that the gradient of $e^{kx}$ is equal to $ke^{kx}$ and hence understand why the exponential model is suitable in many applications